# Apparent power

The apparent power is an operand that in view of the losses must be considered and the stress of the components of a power supply system when an electrical load electric power is supplied. The apparent power does not necessarily match the power passed on by the consumer in the form of thermal, mechanical or other energy. Apparent power  is defined by the effective values of electrical current  and electrical voltage  and is made up of the actual active power and an additional reactive power : ${\ displaystyle S}$${\ displaystyle I}$${\ displaystyle U}$ ${\ displaystyle P}$ ${\ displaystyle Q _ {\ text {dead}}}$

${\ displaystyle S = U \ cdot I = {\ sqrt {P ^ {2} + Q _ {\ text {tot}} ^ {2}}}}$.

All three performance quantities are quantities defined by equivalents or integrals . For them there are no instantaneous values dependent on time for stationary processes . In the case of apparent power defined as unsigned - in contrast to active power - a counting arrow system does not distinguish between consumed and delivered power by means of the sign.

When the reactive power disappears, for example with direct voltage , the apparent power is equal to the amount of the active power, otherwise it is greater. Electrical equipment that is supposed to transmit a specified active power, such as transformers or electrical lines , must be designed for the greater apparent power. The electrical connected load is also often given as apparent power.

Instead of the unit of power watt ( unit symbol  W), the unit volt-ampere (unit symbol VA) is used for apparent power and the unit Var (unit symbol var) for reactive power .

## Apparent power for sinusoidal quantities

In the case of sinusoidal quantities, there is a displacement reactive power when the phase angles of the current intensity and voltage are shifted by one . The voltage and amperage in this case are of the shape ${\ displaystyle Q}$${\ displaystyle \ varphi}$

${\ displaystyle u (t) = {\ sqrt {2}} \; U \ sin \ left ({\ tfrac {2 \ pi t} {T}} \ right)}$
${\ displaystyle i (t) = {\ sqrt {2}} \; I \ sin \ left ({\ tfrac {2 \ pi t} {T}} - \ varphi \ right)}$

In this case, the following applies to the apparent power

${\ displaystyle S = U \ cdot I = {\ sqrt {P ^ {2} + Q ^ {2}}}}$

With     ${\ displaystyle P = U \, I \; \ cos \ varphi}$

and   ${\ displaystyle Q = U \, I \; \ sin \ varphi; \ Q _ {\ mathrm {tot}} = | Q |}$

If an electrical consumer or a supply network contains linear inductances or capacitances , these require electrical energy to build up the magnetic or electrical field, which, however, is returned to the network after every half period . The reactive current required for the field energy is shifted by a quarter period or 90 ° compared to the voltage. The reactive power associated with the transport of the field energy and the active power converted in the consumer result in the Pythagorean sum of the apparent power.

Power vector diagram for sinusoidal quantities

The network and the resources such. B. the supplying generators and transformers must all be rated for the value of the apparent power. This only does not apply if reactive current compensation limits the reactive current to the local consumer-internal lines.

In the complex AC calculation for the sinusoidal voltage or current curve, the apparent power is defined as the amount of the complex apparent power  and as the Pythagorean sum of active power  and reactive power  . The complex apparent power is defined as the product of the complex voltage  and the complex conjugate current  . ${\ displaystyle {\ underline {S}}}$${\ displaystyle P}$${\ displaystyle Q}$${\ displaystyle {\ underline {U}}}$${\ displaystyle {\ underline {I}} ^ {*}}$

${\ displaystyle {\ underline {S}} = {\ underline {U}} \ cdot {\ underline {I}} ^ {*} = P + \ mathrm {j} Q}$
${\ displaystyle S = | {\ underline {S}} | = {\ sqrt {P ^ {2} + Q ^ {2}}}}$

## Apparent power for non-sinusoidal quantities

### The general case

In an electrical network with distorted, ie non-sinusoidal voltages or currents, harmonics occur. Every periodic signal can be broken down into a series of individual sinusoidal oscillations, so-called spectral components , using Fourier analysis . Using the example of the current strength , this consists of ${\ displaystyle I}$

• the fundamental oscillation with the rms value and the phase shift angle to the voltage with the same frequency${\ displaystyle I_ {1}}$ ${\ displaystyle \ varphi _ {1}}$
• the harmonics with and , and , and etc.${\ displaystyle I_ {2}}$${\ displaystyle \ varphi _ {2}}$${\ displaystyle I_ {3}}$${\ displaystyle \ varphi _ {3}}$${\ displaystyle I_ {4}}$${\ displaystyle \ varphi _ {4}}$

In this case, a can no longer be specified. The power factor takes its place${\ displaystyle \ cos \ varphi}$ ${\ displaystyle \ lambda = {\ tfrac {| P |} {S}} \ ;.}$

Examples in which the formulas for sine quantities cannot be used are:

• Non-linear consumers, operated on a sinusoidal voltage source. These contain, for example, rectifiers like those found in power supplies . Distortions occur which affect the apparent power.
• Magnetic circles with a ferromagnetic core material that shows saturation and hysteresis effects - such as B. coils or transformers that do not behave linearly, especially when overdriven, and distort the current.
• Phase angle control with switching on of the current delayed after each zero crossing. At least in the case of the current, there is a time shift in the basic oscillation and the formation of harmonics.

For further calculation, the temporal progressions of the instantaneous values and / or the frequency spectra must be known. ${\ displaystyle u}$${\ displaystyle i}$

In the time domain
${\ displaystyle S = {\ frac {1} {T}} {\ sqrt {\ int _ {t_ {1}} ^ {t_ {1} + T} u ^ {2} (t) \ \ mathrm {d } t \, \ cdot \, \ int _ {t_ {1}} ^ {t_ {1} + T} i ^ {2} (t) \ \ mathrm {d} t}}}$
${\ displaystyle P = {\ frac {1} {T}} \ int _ {t_ {1}} ^ {t_ {1} + T} u (t) \ cdot i (t) \ \ mathrm {d} t }$
In the frequency domain
${\ displaystyle S = {\ sqrt {\ sum _ {k = 1} ^ {\ infty} U_ {k} ^ {2} \, \ cdot \, \ sum _ {l = 1} ^ {\ infty} I_ {l} ^ {2}}}}$
${\ displaystyle P = \ sum _ {k = 1} ^ {\ infty} U_ {k} \, I_ {k} \ \ cos \ varphi _ {k}}$

What contribution the reactive power makes to the apparent power cannot be stated. Just the conclusion about

${\ displaystyle Q _ {\ mathrm {tot}} = {\ sqrt {S ^ {2} -P ^ {2}}}}$

is possible.

### A special case

The tension is often referred to as voltage impressed despite non-linear load without distortion, so  . Then the equations simplify to ${\ displaystyle U = U_ {1}}$

${\ displaystyle S = U \, {\ sqrt {I_ {1} ^ {2} + I_ {2} ^ {2} + I_ {3} ^ {2} + \, \ cdots}}}$
${\ displaystyle P = UI_ {1} \, \ cos \ varphi _ {1}}$

In this case, the reactive power can be specified as consisting of two parts (see also reactive power )

${\ displaystyle Q _ {\ mathrm {tot}} = {\ sqrt {Q_ {1} ^ {2} + Q _ {\ mathrm {d}} ^ {2}}}}$

with a fundamental shift reactive power

${\ displaystyle Q_ {1} = UI_ {1} \, \ sin \ varphi _ {1}}$

and a distortion reactive power caused by the harmonics

${\ displaystyle Q _ {\ mathrm {d}} = U \, {\ sqrt {I_ {2} ^ {2} + I_ {3} ^ {2} + \, \ cdots}}}$

## Problems with switches

### Example dimmer

A circuit consists of a source with sinusoidal voltage, a dimmer and an ohmic load . Here must be considered separately

1. the line between dimmer and consumer (the dimmer is thought to be added to the source) and
2. the line between the source and the dimmer (the dimmer is thought to be added to the consumer).

At the ohmic resistance , every instantaneous value is proportional to${\ displaystyle R}$${\ displaystyle u}$${\ displaystyle i}$

${\ displaystyle u = R \ cdot i}$

The current flows from the "ignition", ie delayed to the zero crossing, until the next zero crossing and accordingly in the second half period. Inserted into the equations for the time domain one comes up ${\ displaystyle \ alpha T}$

${\ displaystyle S = R \ cdot {\ frac {2} {T}} \ int _ {\ alpha T} ^ {T / 2} i ^ {2} \ \ mathrm {d} t}$

and

${\ displaystyle P = R \ cdot {\ frac {2} {T}} \ int _ {\ alpha T} ^ {T / 2} i ^ {2} \ \ mathrm {d} t}$

So here is and there is no distortion reactive power despite the distorted current. The same result can be obtained if one takes into account that there is no phase shift in the ohmic consumer, i.e. that the equations are in the frequency range for the fundamental and all harmonics. ${\ displaystyle S = P}$${\ displaystyle \ varphi _ {k} = 0}$

It is different on the line between the source and the dimmer: the same “dimmed” current flows here, but the voltage is undimmed and sinusoidal. This means that the voltage has a higher rms value and a higher apparent power is generated with unchanged active power. This increase is explained as reactive power, which includes both displacement reactive power and distortion reactive power. However, the displacement reactive power cannot be interpreted as an indication of feedback, because there is no storing component in this example. The more distorted the current, the greater it becomes : with increasing delay in the ignition point in the dimmer, it becomes smaller and smaller, without - until  - the peak value of the current intensity decreases at the same time . ${\ displaystyle S / P}$${\ displaystyle P}$${\ displaystyle \ alpha = 1/4}$

### Example half-wave rectifier

A half-wave rectifier has a similar function when it is used to reduce power, for example in a coffee machine. The rectifier interrupts the energy supply for half a period, i.e. halves the power. The heating plate behaves like an ohmic resistor . A fundamental current with a reduced amplitude and unchanged in phase, plus direct current and harmonic currents, are taken from the source of a sinusoidal alternating voltage. Compared to operation without a rectifier, which is referred to here as the nominal state, this results from the heating plate ${\ displaystyle R}$

${\ displaystyle P = {\ frac {1} {2}} \, P _ {\ mathrm {nominal}} = {\ frac {1} {2}} \; U _ {\ mathrm {nominal}} \ cdot I_ { \ mathrm {Nenn}} = {\ frac {1} {2}} \; R \ cdot I _ {\ mathrm {Nenn}} ^ {2}}$

and at the socket

${\ displaystyle I = {\ frac {1} {\ sqrt {2}}} \, I _ {\ mathrm {nominal}}}$
${\ displaystyle S = U _ {\ mathrm {nominal}} \ cdot I = {\ frac {1} {\ sqrt {2}}} \, P _ {\ mathrm {nominal}}}$.

Since the fundamental oscillation does not experience a phase shift, is . ${\ displaystyle Q = 0}$

Statements cannot be made from the above calculation because of the direct current component in the apparent power. For a suitable solution, see under distortion reactive power . ${\ displaystyle Q _ {\ mathrm {d}}}$

Note: Since this half-wave rectification impresses a direct current component on the load current, this form of power reduction is only permissible for small powers. Otherwise, the upstream local network transformer could be premagnetized and, in the worst case, saturate .